help-39

[rough forge]

after dot product 5cos^2(t)

[regal sequoia]

Yeah but we have 4+3y^2+z^2 in the denominator

What happened to that

[rough forge]

here

[regal sequoia]

Was that not for ln

[gleaming moss]

[ \vec{F} \cdot \psi'(t) = -\sin t \ln (5+\sin^2 t) + 5\cos^2 t +\frac{6\sin t \cos^2 t} {5 + \sin^2 t} + \frac{ 8\sin^3 t \cos^2 t - 8\sin t\cos^4 t}{ 5 + \sin^2 t} ]

[jolly parrot]

shsgd

[regal sequoia]

😭

[rough forge]

it's all the same

4 + 3y² + z²

[regal sequoia]

Ohh

[rough forge]

look

[regal sequoia]

Okay

Okay now I get it

[rough forge]

We are having

\begin{align*} \int_{\psi} &-\sin(t)\ln(5+\sin^2(t))
\ & + 5\cos^2(t) + \frac{6\sin(t)\cos^2(t)}{5 + \sin^2t}
\ & + \frac{-8\sin(t)\cos^2(t)\cdot \left ( \cos^2(t) - \sin^2(t) \right )}{5 + \sin^2(t)} : \dd t
\end{align*}

[jolly parrot]

𝔸dωn𝓲²s

[gleaming moss]

[ \vec{F} \cdot \psi'(t) = -\sin t \ln (5+\sin^2 t) + \frac {25\cos^2 t + 5\cos^2 t\sin^2 t + 6\sin t \cos^2 t + 8\sin^3 t \cos^2 t - 8\sin t\cos^4 t}{ 5 + \sin^2 t} ]

[jolly parrot]

shsgd

[rough forge]

What are the bounds actually? 0 to 1?

oh no

2pi

0 to 2pi

[gleaming moss]

looks like potential to factorise the numerator

and hope it simplifies

[rough forge]

do you have already a bachelors degree in math?

[gleaming moss]

nearly

My final exam is on Wednesday and i'm honestly fucked for it

[rough forge]

whyy

you are very good

[gleaming moss]

It's a Relativity exam, and my lecturer refuses to release past paper solutions so I don't know if i'm doing questions right or not

[rough forge]

damn

that's horrible

maybe you can post them here in a help channel

[gleaming moss]

I tried, nobody replies

[regal sequoia]

Have you called the helpers?

[gleaming moss]

yeah... still no replies

honestly, general relativity is an absolute mindfuck

[rough forge]

wolfram calculated 15.708

[regal sequoia]

What can we factor out

5?

[rough forge]

,w Integrate[(-sin(t)ln(5+sin(t)^2) + 5cos(t)^2+6sin(t)cos(t)^2/(5+sin(t)^2)-8sin(t)cos(t)^2(cos(t)^2-sin(t)^2)/(5+sin(t)^2)),{t,0,2pi}]

[jolly parrot]

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[regal sequoia]

That can’t be right

[rough forge]

hmm idk

https://tenor.com/view/tupac-smoking-smoke-2pac-gif-1812230915449433952

[gleaming moss]

,w dot { ln(5 + sin^2 t), 5cos t + (6sin t cos t / (5+sin^2 t)), (2cos t (cos^2 t - sin^2 t) / (5+sin^2 t))} {-sin t, cos t, -4sin t cos t}

[jolly parrot]

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𝔸dωn𝓲²s

[rough forge]

😂

Troublesome Integral man

There is also the point (0,0,5) mentioned

[gleaming moss]

,w integrate [-sin t ln(sin^2 t + 5) - {8sin t cos^2 t ( cos^2 t - sin^2 t ) / {sin^2 t + 5}} + cos t (5cost + {6sin t cos t / {sin^2t + 5}}), {t,0,2pi}]

[rough forge]

I think Latex doesnt work

with wolfram

,w Intergrate[... , {variable,a,b}]

[jolly parrot]

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[rough forge]

yea

[gleaming moss]

that has to be the answer

[rough forge]

Why is it wrong yummy

[regal sequoia]

That’s weird

[rough forge]

And how the hell you gonna evaluate that

[regal sequoia]

We have never gotten a decimal number before

[rough forge]

haha

[regal sequoia]

On these questions

[rough forge]

I think that's on purpose

to have at least a direction whether it's right or wrong probably

[regal sequoia]

He always does it so that we can calculate it by hand

[gleaming moss]

what is F again?

[rough forge]

I never thought pringles are that complicated

maybe that's why they are expensive

[regal sequoia]

Haha

[gleaming moss]

,w simplify 5x + (6xy / ( 4+3y^2 + z^2) )

[jolly parrot]

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[gleaming moss]

that doesn't help much

[regal sequoia]

No

[rough forge]

at least the circulation is positive

because of the orientation

right?

[gleaming moss]

yeah

i see the problem

z^2 would be (cos^2 t - sin^2 t)^2, so the inside of ln doesn't simplify the same way

,w simplify 4 + 3sin^2 t + (cos^2 t - sin^2 t)^2

[rough forge]

XD

Click on more

[gleaming moss]

,w integrate [-sin t ln(4 + 3sin^2 t + (cos^2 t - sin^2 t)^2) - {8sin t cos^2 t ( cos^2 t - sin^2 t ) / {4 + 3sin^2 t + (cos^2 t - sin^2 t)^2}} + cos t (5cost + {6sin t cos t / {4 + 3sin^2 t + (cos^2 t - sin^2 t)^2}}), {t,0,2pi}]

[jolly parrot]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
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[rough forge]

yes

does that help though haha

[regal sequoia]

Well

This question sucks

[gleaming moss]

,w integrate [-sin t ln( 4 + 3sin^2 t + (cos^2 t - sin^2 t)^2 ) - (8sin t cos^2 t ( cos^2 t - sin^2 t ) / (4 + 3sin^2 t + (cos^2 t - sin^2 t)^2)) + cos t (5cost + {6sin t cos t / {sin^2t + 5}}), {t,0,2pi}]

[rough forge]

[jolly parrot]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by shsgd

[gleaming moss]

why doesn't wolfram understand that

[rough forge]

maybe wolfram is depressed

if you click on the link

it says it doesnt find a result

@gleaming moss

[regal sequoia]

I am so done with this question

[rough forge]

1.34??

[gleaming moss]

that's one hell of a misinterpretation

[rough forge]

maybe it gives up here

after some time

[regal sequoia]

It’s like me

[rough forge]

nuh uh

[regal sequoia]

Yuh hu

[rough forge]

there you go Mrs. Decimals

It's exceeded the amount of characters LOL

[gleaming moss]

there must be some clever sub to do

[regal sequoia]

yeah

[rough forge]

yeahn't

Maybe the surface way wasnt bad after all but we did some mistake

I can imagine that the cross product of gradient and F would resolve

to something easy

because of taking the partials

🚬

[regal sequoia]

yeah maybe

[rough forge]

math was never for us

[lean saddle]

i will have a look at this later ||generally this type of thing is what i love but not at home rn|| but in general when i doubt in this type of this

since sometimes it can be hard to see the patter

wierstass sub

[regal sequoia]

Okay thank you

[lean saddle]

and deal with polinomuials

[regal sequoia]

Yeah

[lean saddle]

imma call them I_1 + I_2 + I_3 + I_4

I_2 is trivial

pick your fave method and use it

for I_1 use sin(t) = u

then you get int u ln(5 + u^2) / sqrt(1-u^2)

[gleaming moss]

before you continue

that's not the correct integral

it should be

[lean saddle]

:(

damn

[gleaming moss]

give me a sec

[lean saddle]

nws

if you could just list them one by one

[gleaming moss]

[ \int_0^{2\pi} -\sin (t) \ln (8 - 7\cos^2 (t) + 4\cos^4 (t)) \dd t ]
[ + \int_0^{2\pi} 5\cos^2 t + \frac{6\sin (t) \cos^2 (t)}{8 - 7\cos^2 (t) + 4\cos^4 (t)} \dd t ]
[ + \int_0^{2\pi} \frac{8\sin^3 (t) \cos^2 (t) - 8 \sin (t) \cos^4 (t)} { 8 - 7\cos^2 (t) + 4\cos^4 (t)} \dd t ]

[lean saddle]

imma ignore the cos^2(t) integral

[jolly parrot]

shsgd

[gleaming moss]

I'm gonna double check that these are the correct integrals, but they should be

[lean saddle]

they good?

ok

so for I_!

I_1

first cos(t) = x

[gleaming moss]

hang on

[lean saddle]

ok

[gleaming moss]

first couple are right

nevermind those are the ones

[rough forge]

THIS IS ACTUALLY (0,0,5)

I think we need to go the surface route

[lean saddle]

should i continue with the sol?

[gleaming moss]

yes I would like to see it

[rough forge]

yea

[lean saddle]

ok

so first cos(t) = x

then you get int ln(8-7x^2 + 4x^4)

then do by parts

which gives you x ln(8-7x^2 + 4x^4) - int ( x (4x^3 - 14x )/(x^4 - 7x^2 + 8)

only dealing with the integral now

first factor out a 2 from the numerator

fuck

i dropped a 4

very early on

one min

[gleaming moss]

ping me when you finish sol

[lean saddle]

ok gimme two min

ill just write up the final answer here

steps would take a while to get through

[rough forge]

with your paremetrization the normal vector should be 0

[gleaming moss]

which param

[rough forge]

psi

[lean saddle]

$$ x\cdot(\ln(4x^4-7x^2+8)-4)+$$

$$\frac{\sqrt{2^\frac{7}{2}+7}(\ln(x\cdot(2x+\sqrt{2^\frac{7}{2}+7})+2^\frac{3}{2})-\ln(x\cdot(2x-\sqrt{2^\frac{7}{2}+7})+2^\frac{3}{2}))+2\sqrt{2^\frac{7}{2}-7}(\arctan(\frac{4x+\sqrt{2^\frac{7}{2}+7}}{\sqrt{2^\frac{7}{2}-7}})+\arctan(\frac{4x-\sqrt{2^\frac{7}{2}+7}}{\sqrt{2^\frac{7}{2}-7}}\right))}{4}$$

for I_1

[gleaming moss]

you can't use psi to eval a surface integral bc psi is a parametrised line

[rough forge]

but how would paremetrize the surface

[jolly parrot]

_MAN_OF_FIRE_
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rough forge]

it's a circle and a some

ok normal vector cant be really 0

but there's gotta be a way now to help with the surface parametrization

if grad x F suggests (0,0,5) then that's gotta be the way

[sick axle]

There are 2n socks, making n pairs of socks with different colors, in a drawer. Randomly select two socks, the probability of having two socks of the same color is p1, the probability of having two socks of different colors is p2. What is the smallest n, such that p2 > 5p1?

please help

oh srry

[lean saddle]

I_2 is

$$\frac{\sqrt{\sqrt{17}+7}(\ln(|2x-\sqrt{2}\sqrt{\sqrt{17}+7}|)-\ln(|2x+\sqrt{2}\sqrt{\sqrt{17}+7}|))+\sqrt{7-\sqrt{17}}\left(\ln(|2x+\sqrt{2}\sqrt{7-\sqrt{17}}|)-\ln(|2x-\sqrt{2}\sqrt{7-\sqrt{17}}|))}{2^\frac{3}{2}\sqrt{17}}$$

[jolly parrot]

_MAN_OF_FIRE_
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lean saddle]

ok so

i dont know how to make it fit

overleaf or smt might help

[gleaming moss]

[ \frac{\sqrt{\sqrt{17}+7}(\ln \abs{ 2x-\sqrt{2}\sqrt{\sqrt{17}+7} } ]
[ -\ln(|2x+\sqrt{2}\sqrt{\sqrt{17}+7}|)) ]
[ +\sqrt{7-\sqrt{17}}\left(\ln(|2x+\sqrt{2}\sqrt{7-\sqrt{17}}|)]
[-\ln(|2x-\sqrt{2}\sqrt{7-\sqrt{17}}|))}{2^\frac{3}{2}\sqrt{17}}]

[jolly parrot]

shsgd
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gleaming moss]

hard to tell what this should look like

[lean saddle]

ok judging by this and the fact the integals are all solvable but ugly

liek this has all just been partial fractions

i dont think this is the right method

like i can solve I_4 if youd like

[regal sequoia]

I think this is unsolvable haha

[lean saddle]

but i dont think this is it

[lean saddle]

its really not

I 4 splits same way as all of these

you factor out a sine

make sin^2 into 1-cos^2

cos sub

then partial fratios

but yk

not pretty

[regal sequoia]

oh okay

[rough forge]

the dot product

[lean saddle]

are we sure sure that there are no issues in the set up?

[rough forge]

with (0,0,5) and the normal vector will get us something nice

[regal sequoia]

no my teacher is known to make mistakes

[lean saddle]

;-;

youre pre uni?

[regal sequoia]

No uni

[gleaming moss]

i'll relieve you of this misery and say no

[gleaming moss]

.

[lean saddle]

i ussually love trig integrals

but this

isnt pretty

this is

inhuman

[regal sequoia]

Yeah I know💀

This is easy according to my teacher

[lean saddle]

an integral i made up that is very very very very pretty is this

everything works perfectly tgt

give this to them

[regal sequoia]

How do we use this?

[brave schooner]

if the gradient is (0,0,5) and the param is $(\cos(\theta), \sin(\theta), r^2(\cos^2(\theta)-sin^2(\theta)))$

[jolly parrot]

caspar

[lean saddle]

last term is cos(2 theta) if that helps?

[rough forge]

yea somthing like that

[brave schooner]

should be solvable pretty nicely

[rough forge]

yea

[rough forge]

i mean what is this 😭

[brave schooner]

theta from 0 to 2pi

[regal sequoia]

Yes

[lean saddle]

i can try to solve it if you set it up

[brave schooner]

r from 0 to 1

[rough forge]

yea

[lean saddle]

numerator fixes itself with a small sub and noticing the 'form'

[regal sequoia]

This is my progress

[lean saddle]

trig terms are -cos(2theta)

[rough forge]

isn't it cos^2 - sin^2

x^2 - y^2 = z

[gleaming moss]

I don't think the curl of F is (0, 0, 5)

[brave schooner]

pretty sure it is

[gleaming moss]

ok, I stand corrected

[lean saddle]

whats the website called?

[rough forge]

https://www.geogebra.org/m/jWfTBWWT

I also wonder

[rough forge]

the normal vector should be some cross product

of two tangent vectors

[brave schooner]

oops yeah

[rough forge]

could the answer be 0?

would that make sense

[brave schooner]

i got that

[rough forge]

the circulation 0 haha (i have no idea)

[brave schooner]

but i had the wrong method

working out the $\hat{N}$ now

[jolly parrot]

caspar

caspar

[rough forge]

jag vet

[brave schooner]

normal vector is (-,-,r)

[brave schooner]

thats soo much easier

[rough forge]

did I do some mistkae?

[brave schooner]

yes, should be rcos(t),rsin(t)

on the second

dont differentiate the cos and sin there

[rough forge]

yea i did

YES

THAT'S IT

[rough forge]

this

so over all

5r

[brave schooner]

yup

[rough forge]

I TOLD YA

[gleaming moss]

can't believe the surface integral route was easier than line integral

good job

[regal sequoia]

Yeah that’s insane

[brave schooner]

so 5pi overall

[regal sequoia]

You were right from the start

[brave schooner]

well generally you will get way easier integrals, so if it doesnt simplify just try again

especially if it tells you to use stokes or gauss

[rough forge]

I was skeptical

[regal sequoia]

Yeah okay

[rough forge]

How the two components get to 0 and 5

and the first is somehow some stupid shit haha

[brave schooner]

i have what seems like a similar class and there are so many problems that simplify like this one

[regal sequoia]

Yeah I hope I never see this again haha

[brave schooner]

well it was not that bad once we got the right curl and normal vector

[rough forge]

yes

[regal sequoia]

Yeah and it’s good to learn if I get one similar to this

[brave schooner]

super common problem

[regal sequoia]

Well thank you very much to all five people who help solve this haha

[brave schooner]

did you finish the divergence problem from last night?

[regal sequoia]

Hmm no

Not yet haha

But I think I will tomorrow

This question took all my brainpower

[brave schooner]

fair

[regal sequoia]

.solved

[pearl pond]

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[pearl pond]

@chrome flame Has your question been resolved?

[pearl pond]

@chrome flame Has your question been resolved?

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@chrome flame Has your question been resolved?

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[indigo plinth]

<@&286206848099549185>

[cunning comet]

pinging without a question. come on.

[indigo plinth]

my wifi sucks bro 😭

[obtuse tendon]

can u translate question?

[cunning comet]

!15min

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[wind condor]

t'es Francais?

[indigo plinth]

no je suis marocain

[wind condor]

I'm Cameroonian

so i understand a bit

let's see

[indigo plinth]

there is a problem with the phrasing of the question

which is why i cant answer it

[pearl pond]

@indigo plinth Has your question been resolved?

[indigo plinth]

@hallow sable

@real tiger

[pearl pond]

@indigo plinth Has your question been resolved?

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[pearl pond]

@scarlet dawn Has your question been resolved?

[quiet tendon]

is this asking you to find the limit using the epsilon-delta definition?

[scarlet dawn]

Yes

[quiet tendon]

okay, well the definition is that $\forall

shit

$\forall \epsilon > 0, \exists \delta > 0, |x - a| < \delta \implies |f(x) - L| < \epsilon$

[jolly parrot]

neil

[quiet tendon]

i don't feel like typing in latex anymore lol so im just gonna type it out

you want to show that if you let epsilon and delta be the same, |x - a| < delta must imply that |sqrt(x^2 + 5) - sqrt(a^2 + 5)| < epsilon

hmm there's usually some trick for manipulating these expressions

oh you can see that sqrt(x^2 + 5) - sqrt(a^2 + 5) is always positive

by showing that x^2 + 5 > a^2 + 5

which means showing that x^2 > a^2

hmm there's gotta be some trick here that i'm not seeing, i'm sorry for wasting your time

[pearl pond]

@scarlet dawn Has your question been resolved?

[scarlet dawn]

What do you think 'bout this proof?

[autumn topaz]

how did you do the second to last step

[strong marsh]

preguntale al ayudante

[hollow void]

There's a shortcut to that problem, you could prove f is Lipschitz with L=1, then you can take \delta = \varepsilon/L.

[pearl pond]

@scarlet dawn Has your question been resolved?

[pearl pond]

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[wet totem]

Did I do this right

[wet totem]

Question b

[pearl pond]

@wet totem Has your question been resolved?

[pearl pond]

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[brave lynx]

I need help setting up a function for this as it goes from -1 to 7

[fresh mural]

you can get 4 values with known outputs and then get a cubic polynomial from there

for example if f(x) is the function graphed, f(-2) = 0, so it must have (x+2) as a factor

[pearl pond]

@brave lynx Has your question been resolved?

[brave lynx]

Okay so

I figured out 7-(-1)/4 is 2

And the squares are weird

[pearl pond]

@brave lynx Has your question been resolved?

[pearl pond]

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[stoic pecan]

i dont understand

their working

n^3/4 a valid test

since 1/n^3/4 isnt even convergent

[tall flint]

it being divergent is the point

[stoic pecan]

oh

bruh

i read answer key wrong

i thought they meant that is covnergent

why do they equate it to n^3/4 anyway

is ln(n)^2 comparable to x^0.5 in the rate it grows?

[tall flint]

equate what?

[stoic pecan]

(ln(n))^2 * sqrt(n) to n^0.75

[tall flint]

they didn't

[stoic pecan]

but they put it over it

[tall flint]

that's completely different from them equating anything

[stoic pecan]

oh

also does n^0.01 /(ln n)^2 still become infinity

[tall flint]

yes

[stoic pecan]

oh

is ln (n) that weak

even like ln(n)^10

[tall flint]

any log in the denominator will eventually get overpowered

it grows super slowly

[stoic pecan]

i see

why is answer key

like

not include 3+x^2

for the first integral

but just 2pi x

[pearl pond]

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[mental verge]

can someone tell me why cosec 856 = cosec 35 and not -cosec35?

I find negative while the answer is positive

[mental verge]

My work

[sterile turtle]

what quadrant is the angle in

[mental verge]

2 i think

[sterile turtle]

yep

and what quads are csc positive in

[mental verge]

where sin is positive like 1 and 4

[sterile turtle]

ooooo

that’s cosine

[mental verge]

oh sorry

yeah 1 and 2 is where sin is positive

[sterile turtle]

you got it now?

[mental verge]

oh yeah i was incorrect at the first place by doing it as -csc145 right?

[sterile turtle]

yea it’s not negative at all

[mental verge]

alright thanks

[pearl pond]

@mental verge Has your question been resolved?

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[somber magnet]

Working on homogenous higher order diffeq. They want us to say if these are dependent or independent. To be dependent they must be scalar multiples, however I dont see how these are scalar multiples?

[somber magnet]

nvm figured it out

.close

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[sick breach]

C)

3rd c)

Here's my working

[rose ibex]

are u solving for equation of c = 0?

cuz the question just gives an expression

[sick breach]

No the other c)

[rose ibex]

ohh

[sick breach]

2sin^2x+cosx=1

[rose ibex]

.rotate

,rotate

[sick breach]

It's wrong

[rose ibex]

the expression u wrote is wrong

[sick breach]

The answers x= 0 2pi/3 4pi/3 2pi

[rose ibex]

its 2(1-cos^2(x)) + cos(x) = 1

u did -cosx

[sick breach]

Oh yh

im still lost

[rose ibex]

show ur updated working

[sick breach]

i got the wrong values

[rose ibex]

i cant really read ur factoring method but it probably went wrong there

just use quadratic formula

the -b+-√b^2... one

[sick breach]

hmm ok

[pearl pond]

@sick breach Has your question been resolved?

[pearl pond]

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[pearl pond]

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[gritty escarp]

I got part A because the three arcs should total 360°, so x = 15°.
I can't figure out part B for the life of me.

[modest badge]

Well if the 3 arc lengths are the same then you'd have an equilateral triangle

For an isosceles, the arc length |PQ| has to equal the arc length of |PR|

[gritty escarp]

Oh shit, that's easy (because we have arc lengths that begins and end at a points on the triangle, right?)
Thanks so much!
So this is scalene because none of the arc lengths are equal.

[modest badge]

should be that way yep

[pearl pond]

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[gritty escarp]

Thanks so much! I really appreciate it!

[pearl pond]

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[stoic pecan]

@autumn fossil you still here/

[stoic pecan]

?

i did the uh thing

and i think i got it?

1/(1-x)^2 = nx^(n-1)

but you can tranform x into the form of 1/x

and make it go down

i got the answer so im probadly right

but not sure

[autumn fossil]

perfect

[stoic pecan]

thanks

[autumn fossil]

note that n/2^(n-1) = n*(1/2)^(n-1)

[stoic pecan]

thats crazy how you can just differntiate it and it works

[autumn fossil]

so x = 1/2

[stoic pecan]

ye

did that

alr tysm man

have a great day

[autumn fossil]

np

[stoic pecan]

.close

[pearl pond]

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[autumn fossil]

you too

[pearl pond]

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[random dirge]

(1.25)^n-1 = .004768716

[random dirge]

Do i use logs like this

Log(1.25)*n-1=log(.0047683716)

Thank

.closw

Z

Z

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[marble blaze]

HELP

[marble blaze]

this is my working out so far

this is the answer, I have a problem understanding the right part

the Find zeros part

[leaden wadi]

Do you know the Quadratic Formula?

[marble blaze]

average rate of change?

[leaden wadi]

x = (-b +- sqrt(b^2 - 4ac)/2a

[marble blaze]

oh yes i know this

formula

can i do it in a calculator

?

[leaden wadi]

The derivative of this equation is a quadratic which you can use QF to find the zeroes if there are any.

[leaden wadi]

Yes.

[marble blaze]

I just have a question why did he use the value -1 and 0

-1 is the average velocity

x(2)-x(0) / 2-0 it leads to an answer of -1

[leaden wadi]

From the first line to the second line on the right, your teacher just added 1 to both sides to the equation on the second line.

[marble blaze]

i dont understnad

can you elaborate

[leaden wadi]

As far as why your teacher set it equal to -1, that allows you to find the values of t for when the expression on the left is equal to -1.

[marble blaze]

im confuesd

the equation is 3t^2-6t+1=-1

we need to solve for t right?

[leaden wadi]

Yes.

[marble blaze]

ok i figured it out

can you help with this

how can i get a second derivative from a table?

[leaden wadi]

Find the slope between the points on both sides and take an average.

[leaden wadi]

,rotate ccw

[leaden wadi]

Looks good.

[marble blaze]

yep its right

last question

[leaden wadi]

If a derivative is positive, what does that imply about a function?

[marble blaze]

its continious?

[leaden wadi]

No. It means the function is increasing.

If g'(x) > 2 on the interval from 1 <= x <= 5, that means it is increasing on that entire interval.

[marble blaze]

so I is wrong?

[leaden wadi]

Why would it be wrong?

[marble blaze]

all of them are true because all of the increased?

[leaden wadi]

If at x=4, g(4) is equal to 3, what are possible values of g(x) before x=4?

[leaden wadi]

While true, all of the values are increasing on the interval, does the increase satisify the constraint that g'(x) > 2?

[marble blaze]

yes

i think so

oh wait

I and II are less than 2

[leaden wadi]

Are you sure about that?

[marble blaze]

Yes

[leaden wadi]

What is a possible slope from g(1) to g(2)?

[marble blaze]

it would be an answer less than 2 right?

[leaden wadi]

Find the slope from g(1) to g(2).

[marble blaze]

there is no table given

[leaden wadi]

Use those values.

[marble blaze]

0-(-6) / 2-1 = 6

so true?

[leaden wadi]

Yes, and what about from g(2) to g(4)?

And then g(4) to g(5)?

[marble blaze]

g(4)-g(2)/4-2 = 3/2

right?

[marble blaze]

and g(5)-g(4)/5-4 = 1

[leaden wadi]

g(4) = 3

So yes.

[marble blaze]

g(4)-g(2)/4-2 = 3/2
g(5)-g(4)/5-4 = 1

II is wrong

and III is wrong

because they are less than 2

g(4)-g(1)/4-1 = 3

so i guess I is the correct answer

[leaden wadi]

Sounds reasonable to me.

[marble blaze]

am I doing it the correct way?

[leaden wadi]

To the best of my knowledge, you ruled out possible answers.

[marble blaze]

yes the answer is correct

but i have a question when i worked out |||

i did g(5)-g(4)/5-4

but in the picture its g(4)-g(5)/5-4

[marble blaze]

how

the average rate of change is f(b)-f(a)/b-a

oh ok

tysm

[leaden wadi]

For the last one, the slope will be the same as long as you subtract the pairs in the same order.

[marble blaze]

thank you so much for the help

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

hi

[midnight haven]

how do i do part a and b

i thought maybe the normal reaction force would just be 6g

but its not

oh wait

ive figured out a

how do i do B

[midnight haven]

the large one right?

yes...?

i have no idea

yeah

but how do i do the question

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[livid tiger]

what does this mean?

[livid tiger]

what is the big T?

in the rhs middle of u and A?

[fluid axle]

a typo

[livid tiger]

what is it supossed tobe?

[fluid axle]

transpose

like the lhs really

[livid tiger]

(uA)^T?

oh just like lhs?

[fluid axle]

nah just u^T Av

[livid tiger]

dot product dissapears?

[fluid axle]

ah nvm

[fluid axle]

so they were a bit drunk when they wrote that sentence ig

u . Av = u^T Av

[livid tiger]

its this one

[fluid axle]

is prolly what they wanted to write

[livid tiger]

the same as for these?

[fluid axle]

yes

[livid tiger]

alright thanks a lot

@fluid axle what is the reason behind that u^T dissapears in the last line?

these ones?

ah

its supossed to be a dot product

.close

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[dense iris]

one box is not equal to one motorist

so how does 7-8 equate to 14

For example the 4-6 range has 4 boxes representing it

so one box should reperesent 3/2 of a motorsit no?

[pearl pond]

@dense iris Has your question been resolved?

[blissful quiver]

The question tricks you

Because the table gives the delay but to include all times it is really 3.5 - 6.5 then 6.5 - 8.5 etc

@dense iris

[dense iris]

wait but how do i know how many motorists

are between that range

[blissful quiver]

It is still the numbers given in the table

[merry carbon]

Also e.g. for the 4 - 6 delay bar [which, because of the "to the nearest minute", is really 3.5 - 6.5min] you can work out that from there, that should represent 6 people in total, 3 * 2

[blissful quiver]

But imagine theyve rounded the numbers in the table to 0 dp

[dense iris]

omg

yeah cause it says 4-6

then goes 7-9

ah

yeah i remember they do this weird thing

in interpolation questions

AHHHHH

[merry carbon]

Yeaa, trippy

[dense iris]

ok kl thx

[merry carbon]

That's how they get the 14, that's a height of 7 and a width of 2

[identical arguments for the 10-12min (really 9.5-12.5min) bar and the 16-20 working out to 5]

[dense iris]

.close

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[tepid isle]

is this setup correct?

to rotate about the x axis

im just a bit confused on finding a function for radius, is it just x?

i mean i guess i could just find the volume from 0 to 2 and subtract volume from 0 to 1

but is there a way to represent in one integral

[covert pivot]

$\pi \int_{1}^{2} {(2 \sqrt{x})}^2 \mathrm{d} x$

[jolly parrot]

Potatomonke

[covert pivot]

simplifies to $4 \pi \int_{1}^{2} x \mathrm{d} x$

[jolly parrot]

Potatomonke

[tepid isle]

oh my god im dumb idk why i thought y axis

[covert pivot]

lol

[tepid isle]

like i got it confused somehow

[covert pivot]

dw

[tepid isle]

ok hypothetically if i were to rotate about the y axis what would i have to do

x axis is pretty straight forward i get that

[covert pivot]

you change the function so its in terms of x

so instead of y = whatever(x)

it becomes x = whatever(y)

since you now differentiate in terms of y

[tepid isle]

would shell method work too? thats what i was initially trying to do

[covert pivot]

well maybe

but you are only dealing with a single curve

so you could just do disc

to rotate around y its just

[tepid isle]

x= (y/2)^2

[covert pivot]

y = 2 sqrt(x) becomes x = (y^2)/4

yeah

then you have to find you new limits

since we are differentiating in terms of y instead of x

[tepid isle]

which are now y limits

yea

[covert pivot]

yea

[tepid isle]

2 and 2sqrt2

[covert pivot]

thats basically it

[tepid isle]

so you would do pi integral from 2sqrt2 to 2 of ((y^2)/4))^2

[covert pivot]

mhm

[tepid isle]

ohh ok

cool thanks

[covert pivot]

dy

[tepid isle]

i probably looked so clueless initially

im so bad at proofreading

[covert pivot]

its fine lol everyone makes mistakes

im grinding out a 15k word essay rn lol

i need to finish at least 1k more before i sleep

[tepid isle]

end of semester is so rough

[covert pivot]

literally

[tepid isle]

im glad im done with college writing

[covert pivot]

oop

[tepid isle]

25 page paper killed me

[covert pivot]

i cant imagine college

if i find this difficult lol

[covert pivot]

rip

honestly we got a month for this but i procrastinated cuz of aps

weve alr done like 10k so its honestly just writing 5k more

[tepid isle]

oh yea im glad im done with aps too lol those sucked

i took ap calc i literally learned this stuff in hs but i gotta review it for my calc 2 final

[covert pivot]

ahh

gl

[tepid isle]

thanks you too

thanks for the help

[covert pivot]

np

[tepid isle]

.close

[pearl pond]

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[acoustic pine]

how do I do this and whats the formula

[acoustic pine]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[willow ginkgo]

a) do not spam helpers right after asking your question
b) put in some effort, show what you tried, share your thought process, etc
c) be more patient

[acoustic pine]

alr my b my b

so what i was thinking was volume = length x width x hight but im just having trouble trying to plug in the numbers for that

wait is it

18 x 1.5 x 2.3^3

= 59.32

kind of a guess but idk

<@&286206848099549185>

[elder swan]

Bro dbi why you taggin bers

[acoustic pine]

?

[elder swan]

what’s your question I’ll do just so you shush

[acoustic pine]

this

[tepid isle]

so i think youre just converting in^3 to ft^3

question has awful wording

but im pretty sure thats whats going on

[acoustic pine]

was this right then?

or naw

[tepid isle]

hmm

i think you would multiply the original volume by 12^3

since there are 12 inches in a foot

and its cubed

[acoustic pine]

when you say original volume do you mean 2.3^3 ft

or wdym

[acoustic pine]

you mean this volume?

[tepid isle]

yea the given

im like 90% sure

[acoustic pine]

alr ty

.close

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[acoustic pine]

.reopeb

.reopen

[pearl pond]

✅

[acoustic pine]

just to clarify to find the answer you said to mulitply

2.3^3 x 12^3

[tepid isle]

oh the 2.3 is ft^3

its not like 2.3^3

its just 2.3

and ft^3 is the units

sorry i didnt catch that before

[acoustic pine]

so 2.3 x 12^3?

[tepid isle]

yea

try that

[acoustic pine]

k

ty

the final answer comes out as 3974.4

but I feel like thats kinda long no?

.close

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[echo matrix]

Is the answer -1/4 for both x and y

I am not sure if I am right but this is what I got

[pearl pond]

@white walrus Has your question been resolved?

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[vocal lark]

.reopen

[pearl pond]

✅

[vocal lark]

@white walrus its polite to say thank you to people who help you with your math questions.

[pearl pond]

@white walrus Has your question been resolved?

[pearl pond]

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[coral bone]

Each letter in the word ACRIMONIOUS is printed individually on a card. When cards are arranged next to each other in a line, determine the number of different permutations of all the cards where all the consonants are adjacent

[steel canopy]

wht did u try to do?

[coral bone]

Theres 6 vowels and 5 consonants so there would be two groups of vowels and consonants

[steel canopy]

yes

[coral bone]

So i did (6! x 5! x 2!)/2!2!

[steel canopy]

2! for 2 o's and 2 i's okay

[coral bone]

Yeah

[steel canopy]

its correct

[coral bone]

The answers say its (7!x5!)/2!2!

[steel canopy]

oh ya y did u consider all vowels together

they can be anywhere

[coral bone]

What

[steel canopy]

like 2 to right to consonants and other to left like that

so yea 7! comes

[coral bone]

Im not understanding

[steel canopy]

lemme show u

[coral bone]

Ok

[steel canopy]

c for consonant and v for vowel

[coral bone]

Yea

[steel canopy]

now shffle in 7! ways right

each vowel is separate

[coral bone]

OHH

[steel canopy]

so total 7 characters

[coral bone]

And theres 5 consonants

Ah i see

So its 7 groups

[steel canopy]

yes

[coral bone]

So thats why its 7!

Ok thank you

.close

[pearl pond]

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[coral bone]

.reopen

[pearl pond]

✅

[coral bone]

Wait i have another question

How would you determine the number of different permutations using any 4 of the letters

[steel canopy]

wot?

[coral bone]

So like

You choose 4 letters

[steel canopy]

ok

and then write permutaions?

ACRIMONIOUS

[coral bone]

Well

[steel canopy]

so A,C,R,I,M,O,N,U,S

[coral bone]

Yeah

[steel canopy]

so choose 4

9C4

and shuffle 4!

if thats wht u are asking

[coral bone]

Why is it 4!?

[steel canopy]

because u have a word with 4 letters

[coral bone]

Oh

[steel canopy]

yes

[coral bone]

The answers also added the times where there were pairs of letters and both pairs

[steel canopy]

oh so we need to consider those

ok so now extra 2 letters are there

I and O

getting repeated

[coral bone]

Yea

[steel canopy]

so consider u r getting 2 I's

so 2 letters left

[coral bone]

Mhm

[steel canopy]

choose - 8C2
shuffle - 4!/2!

if u consider 2 O's its same

but if u get 2I's and 2O's

then 4!/2!2!

[coral bone]

Oh thats why its divided by 2!

[steel canopy]

yes

[coral bone]

Huh

How do you know so much

[steel canopy]

idk i love maths

[coral bone]

Do you also know about vectors?

[steel canopy]

yes

[coral bone]

Ok

P and Q are moving with constant velocities (2,0)ms and (1,-2)m/s respectively. P has initial position vector (1,-2)m and Q has initial position vector (3,4). Determine the distance between the vectors after 5 seconds

[steel canopy]

dude this is physics only

[coral bone]

Well my syllabus has vectors that go into physics and forces 😭

[steel canopy]

mhm

fine

[coral bone]

I know about the vector stuff

[steel canopy]

just write it as x and y directions

apply newton laws

[coral bone]

Actualy wait

This is a bit more simpler

You just addon 5 x their speed

[steel canopy]

wot

yes

yes

[coral bone]

Ok i dont think i need anymore help for noe

.close

[pearl pond]

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[pearl pond]

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[woven oak]

s a m e

[woven oak]

hey close this ticket i didnt make question

ok this ig

[steel canopy]

x=cost

[pearl pond]

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[steady crescent]

Trig sub probs your best bet here. Let x=sinx works I think

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[midnight haven]

4th

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[drowsy flare]

My text says e^(10x)=100x+1 has two solutions, but I can not see how it is 2 and not 1, i tried on graph and it looks like 1 there too

[old marsh]

,w graph e^(10x) - 100x-1 on [-0.5,0.5]

[gloomy scroll]

It has two sols because the Lambert W function has multiple branches

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Requested by northsteve

[old marsh]

There we go lol

[pearl pond]

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[drowsy flare]

How to solve without graph?

?

[pearl pond]

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[dim vortex]

for a functional equation like $$f(\frac{x+2}{x-2})=\frac{x^2+4x+4}{8x}$$ how would u ask wolfram alpha to solve for f(x) ?

[jolly parrot]

Hemesfere

[sharp smelt]

let (x+2/x-2)=u

[dim vortex]

ok?

how do i ask tho

[sharp smelt]

change it to an equation in u I guess

[dim vortex]

this is what i got

its not the answer tho

[sharp smelt]

write it in terms of f(y)

you'll have to solve the equation first

to obtain x in terms of y

,w solve (x+2)/(x-2)=u for x

[dim vortex]

thats what im trying to skip lol

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[sharp smelt]

substitute x with this

[dim vortex]

ik how to solve the equation, im asking if theres a way to solve functional equations using wolfram alpha

[sharp smelt]

not sure

[dim vortex]

ok ty👍

.c

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[zinc trellis]

Can vectors just go on each other like that?

I need to do a-b but they kinda end up being on the same line ?????

[arctic breach]

Yes

Free vector

[zinc trellis]

so it's all good?

[arctic breach]

Yes

[zinc trellis]

so is it 0° then?

Or 180

[arctic breach]

Depends on Directoion

*direction

[zinc trellis]

how

[arctic breach]

The pointing arrow

If arrow is on left hand side then 180

If right then 0

[zinc trellis]

ohhhh

got it

thanks !

.c

.close

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[white walrus]

Can someone explain alternative method